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A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by $v(t) = 2t (3 - t); 0<t<3$ and $v(t) = -(t - 3) ( 6 - t)$ for $3 < t < 6s$ on m/s. It repeats this cycle till it reaches the height of 20 m.

a) At what time is its velocity maximum?

b) At what time is its average velocity maximum?

c) At what times is its acceleration maximum in magnitude?

d) How many cycles are required to reach the top?

Answers (1)

(a) for the velocity to be maximum,

$\frac{dv(t)}{dt}=0$

$d\frac{[2t(3-t)]}{dt}=0$

$d\frac{6t-2t^{2}}{dt}=0$

$6 - 4t = 0$

$4t = 6$

Thus, $t=\frac{3}{2}$

$= 1.5 s$

(b)Now, average velocity

$V=\frac{x}{t}$

$=6t-2t^{2}$

$\frac{ds(t)}{dt}=6t-2t^{2}$

$ds=(6t-2t^{2})dt$

Let us integrate both the sides from 0 to 3

$\int_{0}^{s}ds=\int_{0}^{3}(6t-2t^{2})dt$

$s=\left [ 6\frac{t^{2}}{2}-2\times\frac{t^{3}}{3} \right ]_{0}^{3}=\left [ 3t^{2}-\frac{2}{3}t^{3} \right ]_{0}^{3}$

$=\left [ 3\times9-\frac{2}{3}\times2 \right ]=27-18$

Thus, $s = 9m$

Now, average velocity $(V_{av})=\frac{9}{3}=\frac{3m}{s}$

$V(t) = 6t - 2t^{2} .......... since \; 0 < t < 3$

$3 = 6t - 2t^{2} .......... since V_{av}=3$

$2t^{2} - 6t + 3 = 0 \; \; \; \; \; \; .............. since ( a=2, b=-6 \; and\; c=3)$

$t=2.36s$

Thus, the average velocity is maximum at 2.36 seconds.

(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.

$Here, v=0$

$V(t) = 6t - 2t^{2}$

$0 = 6t - 2t^{2}$

$2t (3-t) = 0$

$t\neq 0$

Thus, the acceleration is maximum at t = 3 sec.

(d)Now, for 3 to 6 sec

$V(t) = -(t-3) (6-t)$

$\frac{ds}{dt}=(t-3)(6-t)$

$ds=(t^{2}-9t+18)dt$

Integrate from 3 to 6 s

$s_{2}=\int_{3}^{6}(t^{2}-9t+18)dt=\left [ \frac{t^{3}}{3}-\frac{9}{2}t^{2}+18t \right ]_{3}^{6}$

$=\frac{(6)^{3}}{3}-\frac{9}{2}(6)^{2}+18\times6-\left [ \frac{(3)^{3}}{3}-\frac{9}{2}(3)^{2}+18\times3 \right ]$

$=\frac{6\times6\times6}{3}-\frac{9\times6\times6}{2}+108-\frac{3\times3\times3}{2}-54$

$=180-162-63+40.5=18-22.5$

$S_{2}=-4.5\; m$ .............because distance is in downward direction

Thus, net distance $=4.5\; m$

Thus, in three cycle $=4.5(3)$

$=13.5\; m$

Remaining height will be

$20 -13.5 = 6.5 m$

The monkey can climb up to 9m without slipping but in the 4^thcycle it will slip and the height remaining to climb will be 6.5 m.

Net no. of cycle = 4.

Posted by

infoexpert22

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